如何正确地将unsigned char数组转换为uint32_t [英] How to properly convert an unsigned char array into an uint32_t

问题描述

所以，我试图将 unsigned char 的数组转换为 uint32_t ，但继续获取每次不同的结果：

  unsigned char buffer [] = {0x80,0x00,0x00,0x00} ;; 
 uint32_t num =（uint32_t *）& buffer; 
  

现在，我一直收到此警告：

< blockquote>

警告：初始化从指针无整数转换

当我更改 num 到 * num 我没有得到那个警告，但这不是真正的问题（ UPDATE：好吧，那些可能是相关的，现在我想到它。），因为每次我运行代码有不同的结果。其次， num ，一旦正确转换，应该是 128 ，但如果我需要更改缓冲区的字节顺序

谢谢！

解决方案

您尝试过这个吗？

  num =（uint32_t）buffer [0] < 24 | 
（uint32_t）buffer [1]< 16 | 
（uint32_t）buffer [2]< 8 | 
（uint32_t）buffer [3]; 
  

这样就可以控制字节序和其他。

<指向一个 char 指针并将其解释为更大的东西真的不安全。有些机器需要指向整数的指针对齐。

So, I'm trying to convert an array of unsigned chars into an uint32_t, but keep getting different results each time:

unsigned char buffer[] = {0x80, 0x00, 0x00, 0x00};;
uint32_t num = (uint32_t*)&buffer;

Now, I keep getting this warning:

warning: initialization makes integer from pointer without a cast

When I change num to *num i don't get that warning, but that's not actually the real problem (UPDATE: well, those might be related now that I think of it.), because every time I run the code there is different results. Secondly the num, once it's cast properly, should be 128, but If I need to change the endianness of the buffer I could manage to do that myself, I think.

Thanks!

解决方案

Did you try this ?

num = (uint32_t)buffer[0] << 24 |
      (uint32_t)buffer[1] << 16 |
      (uint32_t)buffer[2] << 8  |
      (uint32_t)buffer[3];

This way you control endianness and whatnot.

It's really not safe to cast a char pointer and interpret it as anything bigger. Some machines expect pointers to integers to be aligned.

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